Valuations on Rational Function Fields That Are Invariant Under Permutation of the Variables

VALUATIONS ON RATIONAL FUNCTION FIELDS THAT ARE INVARIANT UNDER PERMUTATION OF THE VARIABLES

F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

Abstract. We study and characterize the class of valuations on rational functions ﬁelds that are invariant under permutation of the variables and can be extended to valuations with the same property whenever a ﬁnite number of new variables is adjoined. The Gauß valuation is in this class, which constitutes a natural generalization of the concept of Gauß valuation. Further, we ap- ply our characterization to show that the most common ad hoc generalization of the Gauß valuation is also in this class.

1. Introduction In this paper, we will work with (Krull) valuations v and write them in the classical additive way, that is, the value group of v on a ﬁeld K, denoted by vK, is an additively written ordered abelian group, and the ultrametric triangle law reads as v(a+b) ≥ min{va, vb}. We denote by Kv the residue ﬁeld of v on K, by va the value of an element a ∈ K, and by av its residue.

Take any field K and a rational function field K(X1,...,Xm). In [6], the possible extensions of a valuation on K to K(X1,...,Xm) have been studied. While a quite good description of all possible value groups and residue fields of the extensions has been achieved, a detailed description and characterization of the extensions themselves seems rather out of reach. In this situation, restricting the problem to a subset of extensions with additional properties may well be of help. In situations where at least one extension of a nontrivial valuation of K has to be constructed on K(X1,...,Xm), the Gauß valuation is often the canonical choice. Once we have a valuation on

Date: 22. 6. 2016. 2000 Mathematics Subject Classification. Primary 12J10; secondary 13A18. Key words and phrases. Valuations on rational function fields, symmetric valuation, Gauß valuation. The authors would like to thank Anna Blaszczok for her careful reading of the paper and several helpful remarks and corrections. 1 2 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN a polynomial ring, it extends in a unique way to the quotient field. On K[X1,...,Xm], the Gauß valuation is defined as follows. Given f ∈ K[X1,...,Xm], write

X i1 im (1) f = diX1 · ... · Xm i where the sum runs over multi-indices i = (i1, . . . , im) and each di is an element of K. Then deﬁne

(2) vf := min vdi . i

This indeed defines a valuation on K[X1,...,Xm]; see Corollary 2.2 below. We have occasionally witnessed ad hoc attempts to generalize the concept of Gauß valuation, but these attempts did not consider its particular properties in any systematic way. So let us take a closer look at some of them. First, we notice that definition (2) is invariant under permutation of the variables. Second, the same definition can be used to extend the valuation to K[X1,...,Xn] for each n > m, preserving the property of being invariant. Third, it is well known (and follows from Lemma 2.4 below) that for the Gauß valuation v on K[X1,...,Xn], the residues X1v, . . . , Xmv are algebraically independent over the residue field Kv. This implies that v is an Abhyankar valuation on the function field K(X1,...,Xn)|K; the definition of this notion will be given after The- orem 1.3. The property that the residues X1v, . . . , Xmv are algebraically independent over Kv singles out the Gauß valuation. In order to obtain a generalization of the notion of Gauß valuation, we have to drop this property. Our goal therefore is to characterize the valuations that have the first two properties. Thereafter we will clarify their relation to the property of being Abhyankar valuations.

For each permutation π ∈ Sm we denote by τπ the automorphism of K(X1,...,Xm) over K induced by the corresponding permutation

(X1,...,Xm) 7→ (Xπ(1),...,Xπ(m)) .

Its restriction to the polynomial ring K[X1,...,Xm] is an automorphism as well. A valuation v on K(X1,...,Xm) or K[X1,...,Xm] will be called symmetric if vτπ = v for all π ∈ Sm , that is, if it is invariant under permutation of the variables. Several classes of symmetric valuations have been studied in [9, 10, 11, 12]. INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 3

The present paper revisits the content of [10] and presents it in a concise form, with shorter and more accessible proofs. A symmetric valuation v on K(X1,...,Xm) (or K[X1,...,Xm]) will be called openly-symmetric if it can be extended to a symmetric valuation on K(X1,...,Xn) (or K[X1,...,Xn], respectively) for every n > m. Note that such valuations were called “symmetrically-open” in [10], but this name appears to indicate a topological meaning, which is not intended. As we have seen above, Gauß valuations are openly-symmetric. A valuation is symmetric on K(X1,...,Xm) if and only if it is symmetric on K[X1,...,Xm], and the same holds for “openly-symmetric”. Further, every valuation on the polynomial ring K[X1,...,Xm] defines a valuation on the rational function field K(X1,...,Xm). Therefore, as was done in [10], we will often work with the polynomial ring in place of the rational function field. In Section 3 we will prove the following useful result. By Kac we will denote the algebraic closure of K. Proposition 1.1. If v is an openly-symmetric valuation on the polyno- ac mial ring K[X1,...,Xm], then each of its extensions to K [X1,...,Xm] is again openly-symmetric. Remark 1.2. Take a partition I∪˙ J = {1, . . . , m}. If v is symmetric on K[X1,...,Xm], then it is also symmetric on the polynomial ring

K(Xi | i ∈ I)[Xj | j ∈ J]

(with respect to permutations of the variables Xj, j ∈ J). If v is openly-symmetric on K[X1,...,Xm], then it is also openly-symmetric on K(Xi | i ∈ I)[Xj | j ∈ J]. From Proposition 1.1, it follows that ac each extension of v to the polynomial ring K(Xi | i ∈ I) [Xj | j ∈ J] is again openly-symmetric. In Section 5 we will give an example which shows that the openness condition in Proposition 1.1 is necessary. There we will construct a valuation on Qac(X,Y ) which is not symmetric, but has a symmetric restriction to Q(X,Y ). (This example shows that Proposition 3.5 of [9] is in error; its attempted proof contains a gap that cannot be ﬁlled in general. But it can be ﬁlled in the case of openly-symmetric valuations, as Proposition 1.1 shows.) As a consequence of Proposition 1.1, every openly-symmetric valuation on K[X1,...,Xm] is the restriction of an openly-symmetric valua- ac tion on K [X1,...,Xm]. This fact is a very useful tool in the description of the variety of openly-symmetric valuations on K[X1,...,Xm]. 4 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

The following theorem summarizes our main results; it will be proved in Section 4. Note that K[X1,...,Xm] = K[X1,X2 − X1 ...,Xm − X1] and that every polynomial f ∈ K[X1,...,Xm] can be written in the form

X i2 im (3) f = gi(X1)(X2 − X1) · ... · (Xm − X1) i where the sum runs over multi-indices i = (i2, . . . , im) and each gi(X1) is a polynomial in K[X1], Theorem 1.3. Take an arbitrary ﬁeld K and a valuation v on the rational function ac ﬁeld K (X1,...,Xm). a) The valuation v is openly-symmetric on K[X1,...,Xm] if and only ac if for each polynomial f ∈ K [X1,...,Xm] written in the form (3), we have that

(4) vf = min (vgi(X1) + (i2 + ... + im)δ) , i where δ is any value in some ordered abelian group extension of the value group vK(X1) that satisﬁes ac (5) δ ≥ v(X1 − a) for all a ∈ K . b) If v is openly-symmetric on K[X1,...,Xm] and m ≥ 3, then the elements X − X X − X 3 1 ,..., m 1 X2 − X1 X2 − X1 are units in the valuation ring and their residues are algebraically independent over the residue ﬁeld K(X1,X2)v. c) With r := dimQ vK(X1,...,Xm)/vK and t := trdeg K(X1,...,Xm)v|Kv, exactly the following cases are possible: (a) r = 0 and t = m, (b) r = 0 and t = m − 1, (c) r = 1 and t = m − 1, (d) r = 1 and t = m − 2, (e) r = 2 and t = m − 2. Therefore, we always have that m − 1 ≤ r + t ≤ m . Cases (a), (c) and (e) always appear for arbitrary valuations on K, and cases (b) and (d) always appear when the valuation v on Kac admits INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 5

ac an extension to K (X1) that does not enlarge value group and residue field. ac Note that once a valuation v on K (X1) is given, the formula (4) de- ac ac fines a valuation on K [X1,...,Xm] and thus also on K (X1,...,Xm), for any δ (see Corollary 2.2 below). As a consequence of our theorem, arbitrary valuations on K(X1) can be extended to openly-symmetric valuations on K(X1,...,Xm). In the paper [1] (see also [6] and [2]), conditions are discussed under which for a given valued field K its algebraic closure Kac admits extensions of the valuation to the rational ac function field K (X1) that do not enlarge value group and residue field. In the cases (a), (c) and (e) we have that

r + t = trdeg K(X1,...,Xm)|K, that is, v is an Abhyankar valuation of the rational function field K(X1,...,Xm)|K. But the remaining cases show that an openly- symmetric valuation of a rational function field is not necessarily an Abhyankar valuation, so it may not share this particular property with the Gauß valuation, which belongs to case (a). However, openly- symmetric valuations can fail to have this property by a transcendence degree of at most one. Our theorem shows that if v is openly- symmetric, then for each i ∈ {1, . . . , m}, v is an Abhyankar valuation of K(X1,...,Xm)|K(Xi). The importance of Abhyankar valuations can be seen from the results of the papers [5, 7], and they now play a considerable role in algebraic geometry. Finally, let us make a few more remarks about the characterization of openly-symmetric valuations. If we are working over a field K that is not algebraically closed and we only know that condition (4) is satisfied for all polynomials f ∈ K[X1,...,Xm], can we still conclude that v is openly-symmetric on K[X1,...,Xm]? The following corollary to the proof of our main theorem shows that we can, if we strengthen condition (5). Corollary 1.4. Take K and v as in Theorem 1.3. Then v is openly- symmetric on K[X1,...,Xm] if and only if (4) holds for each polynomial f ∈ K[X1,...,Xm], where δ is any value in some ordered abelian group extension of the value group vK(X1) that satisfies ac (6) δ ≥ v(Xi − a) for all a ∈ K and 1 ≤ i ≤ m . The Gauß valuation is an example of a valuation that satisfies conditions (4) and (6). However, when we construct other openly-symmetric valuations, checking condition (6) could be more work than we are 6 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN ready to invest. In this case, the following result is helpful. Again, it follows readily from the proof of Theorem 1.3. Corollary 1.5. Take K and v as in Theorem 1.3. If (4) holds for each polynomial f ∈ K[X1,...,Xm], where δ is any value in some ordered abelian group extension of the value group vK(X1) that satisfies ac (7) δ > v(X1 − a) for all a ∈ K , then v is openly-symmetric on K[X1,...,Xm].

It is not true that a symmetric valuation on K[X1,...,Xm] is openly- symmetric already when it can be extended to a symmetric valuation on K(X1,...,Xm+1). But the latter turns out to be suﬃcient when it is combined with additional conditions in the spirit of Remark 1.2.

ac Corollary 1.6. Take a valuation v on K(X1,...,Xm+1) and assume ac that for 1 ≤ i ≤ m, v is symmetric on K(X1,...,Xi−1) [Xi,...,Xm+1]. Then v is openly-symmetric on K[X1,...,Xm]. To conclude with, we will show how Theorem 1.3 can be applied to analyze the most common ad hoc generalization of the Gauß valuation. Take a rational function ﬁeld K(X1,...,Xm) over the valued ﬁeld (K, v) and an element δ in some ordered abelian group containing vK. For every f ∈ K[X1,...,Xm] written in the form (1), set

(8) vδf := min (vdi + (i1 + ... + im)δ) . i

By Corollary 2.2 below, this deﬁnes a valuation vδ on K(X1,...,Xm) which extends v.

Corollary 1.7. The valuation vδ on K(X1,...,Xm) is openly-symmetric.

2. Some preliminaries For the easy proof of the following lemma, see [3], Chapter VI, §10.3, Theorem 1. Lemma 2.1. Let (L|K, v) be an extension of valued ﬁelds. Take elements xi, yj ∈ L, i ∈ I, j ∈ J, such that the values vxi , i ∈ I, are rationally independent over vK, and the residues yjv, j ∈ J, are algebraically independent over Kv. Then the elements xi, yj, i ∈ I, j ∈ J, are algebraically independent over K. INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 7

Moreover, write

X Y µk,i Y νk,j (9) f = ck xi yj ∈ K[xi, yj | i ∈ I, j ∈ J] k i∈I j∈J in such a way that whenever k 6= `, then there is some i s.t. µk,i 6= µ`,i or some j s.t. νk,j 6= ν`,j . Then

Y µk,i Y νk,j X (10) vf = min v ck xi yj = min vck + µk,ivxi . k k i∈I j∈J i∈I That is, the value of the polynomial f is equal to the least of the values of its monomials. In particular, this implies: M vK(xi, yj | i ∈ I, j ∈ J) = vK ⊕ Zvxi i∈I

K(xi, yj | i ∈ I, j ∈ J)v = Kv (yjv | j ∈ J) .

The valuation v on K(xi, yj | i ∈ I, j ∈ J) is uniquely determined by its restriction to K, the values vxi and the fact that the residues yjv, j ∈ J, are algebraically independent over Kv. The residue map on K(xi, yj | i ∈ I, j ∈ J) is uniquely determined by its restriction to K, the residues yjv, and the fact that values vxi , i ∈ I, are rationally independent over vK.

Conversely, if δi , i ∈ I, are elements of some ordered abelian group containing vK and are rationally independent over vK, then

Y µk,i Y νk,j X (11) vf := min v ck xi yj = min vck + µk,iδi k k i∈I j∈J i∈I deﬁnes a valuation that extends the valuation v from K to the rational function ﬁeld K(xi, yj | i ∈ I, j ∈ J).

Corollary 2.2. Take a rational function field K(X1,...,Xm) over the valued field (K, v) and an element δ in some ordered abelian group containing vK. Then the function vδ defined in (8) induces a valuation on K(X1,...,Xm) which extends v. Proof. We distinguish two cases. First, suppose that δ is a torsion element over vK. Take any extension of v to Kac. Then δ = vb for some b ∈ Kac. In this case we set −1 yi := b Xi for 1 ≤ i ≤ m. By the second part of Lemma 2.1, setting

X i1 im w ciy1 · ... · ym := min vci i i 8 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

ac ac induces a valuation on K (y1, . . . , ym) = K (X1,...,Xm) that extends v. For f written in the form (1) we have that

X i1 im X i1+...+im i1 im wf = w diX1 · ... · Xm = w dib y1 · ... · ym i i

= min (vdi + (i1 + ... + im)δ) = vδf , i which proves that vδ is a valuation. Now suppose that δ is rationally independent over vK. In this case −1 we set x := X1 and yi := x Xi for 2 ≤ i ≤ m. By the second part of Lemma 2.1, setting

X i1 i2 im w cix y2 · ... · ym := min (vci + i1δ) i i induces a valuation on K(x, y2, . . . , ym) = K(X1,...,Xm) that extends v. For f written in the form (1) we have that

X i1 im X i1+...+im i2 im wf = w diX1 · ... · Xm = w dix y2 · ... · ym i i

= min (vdi + (i1 + ... + im)δ) = vδf , i which again proves that vδ is a valuation. Let (L|K, v) be an extension of valued ﬁelds of ﬁnite transcendence degree. Then the following well known form of the “Abhyankar inequality” is a consequence of Lemma 2.1:

(12) trdeg L|K ≥ rr vL/vK + trdeg Lv|Kv , where rr vL/vK := dimQ (vL/vK) ⊗ Q is the rational rank of the abelian group vL/vK, i.e., the maximal number of rationally independent elements in vL/vK. The straightforward proof of the following lemma is left to the reader. Lemma 2.3. The value group of an algebraically closed ﬁeld is divisi- ble, and its residue ﬁeld is algebraically closed.

We will also need the following easy result: Lemma 2.4. Take a valued ﬁeld (K(x), v). If a ∈ K is such that (13) v(x − a) = max{v(x − c) | c ∈ K} , then either v(x − a) ∈/ vK, or for every d ∈ K with vd = v(x − a), we x−a have that d v∈ / Kv. INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 9

If in addition K is algebraically closed, then in the ﬁrst case, v(x−a) x−a is rationally independent over vK, and in the second case, d v is transcendental over Kv. Proof. Assume that v(x − a) ∈ vK and take d ∈ K with vd = v(x − a). x−a x−a Assume further that d v ∈ Kv. Then take c ∈ K such that d v = cv, so that x − a v − c > 0 . d Then v(x − a − cd) > vd = v(x − a), which contradicts (13) since a + cd ∈ K. This proves the ﬁrst assertion of the lemma. The second assertion of the lemma follows by Lemma 2.3 from what we have already shown.

3. Proof of Proposition 1.1 We will need the following auxiliary result. Lemma 3.1. Let m, n, k be natural numbers such that n = m + 2k. Assume that v is a valuation on K[X1,...,Xn] whose restriction to K[X1,...,Xm] is not symmetric. Then the set

{vτπ | π ∈ Sn} has at least k + 1 many elements.

Proof. Since v is not symmetric on K[X1,...,Xm], there is π ∈ Sm such that vτπ 6= v on K[X1,...,Xm]. Since Sm is generated by trans- positions, vτπ 6= v must already hold for some transposition π. W.l.o.g. we may assume that π = (1 2). Pick some f ∈ K[X1,...,Xm] such that vτπf 6= vf. Now using (1 2) as an element of Sn, we ﬁx the automorphism τ = τ(1 2) of K[X1,...,Xn]. Then we have that vτf 6= vf. Pick i, j such that m < i < j ≤ n. Let τi j = τ(1 i)(2 j) be the automorphism of K[X1,...,Xn] that exchanges X1 with Xi and X2 with Xj and leaves the other variables ﬁxed. We have that

fi j := τi jf = f(Xi,Xj,X3,...,Xm) ∈ K[Xi,Xj,X3,...,Xm] and that vτi jfi j = vf 6= vτf = vττi jfi j .

It follows that the restrictions of vτi j and vττi j to the polynomial ring K[Xi,Xj,X3,...,Xm] are distinct, so at least one of them is distinct from the restriction of v to K[Xi,Xj,X3,...,Xm]. If it is vτi j, then we set vi j := vτi j; otherwise, we set vi j := vττi j. 10 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

Take i0, j0 such that m < i0 < j0 ≤ n and {i, j}∩{i0, j0} = ∅. Then the restrictions of τi j and τ to K[Xi0 ,Xj0 ,X3,...,Xm] are the identity and therefore, v and vi j coincide on K[Xi0 ,Xj0 ,X3,...,Xm], which shows that vi j 6= vi0 j0 . By what we have proved, it follows that the k + 1 many valuations

v, vm+1 m+2 , vm+3 m+4 , . . . , vn−1 n are distinct on K[X1,...,Xn].

Now we are ready for the Proof of Proposition 1.1:

Take a valuation v on K[X1,...,Xm] which is openly-symmetric. Fur- ac ther, choose some extension to K [X1,...,Xm] and call it again v. ac Suppose that v is not symmetric on K [X1,...,Xm]. Then there is already a ﬁnite normal extension L|K so that the restriction v0 of v to L[X1,...,Xm] is not symmetric. Set k = [L : K] and n = m + 2k. Then choose a symmetric extension w of v from K[X1,...,Xm] to K[X1,...,Xn]. We show that there is an extension of v0 to L[X1,...,Xn] which 0 also extends w. Choose any extension w of w to L[X1,...,Xn]. The 0 restriction w0 of w to L[X1,...,Xm] is like v0 an extension of v from K[X1,...,Xm] to L[X1,...,Xm]. As the extension

L[X1,...,Xm]|K[X1,...,Xm] is algebraic and normal, v0 and w0 are conjugate over K[X1,...,Xm] (cf. [4, Theorem 3.2.15]). Choose an automorphism σ0 of L[X1,...,Xm] over K[X1,...,Xm] such that v0 = w0σ0. The purely transcendental extension

K[X1,...,Xn]|K[X1,...,Xm] is linearly disjoint from the algebraic extension

L[X1,...,Xm]|K[X1,...,Xm]

(cf. [8, Chapter X, $5, Prop. 3]). Therefore, σ0 can be extended to an automorphism σ of L[X1,...,Xn] over K[X1,...,Xn]. The re- 0 striction of w σ to L[X1,...,Xm] is w0σ0 = v0. Since σ is trivial on 0 K[X1,...,Xn], the restriction of w σ to K[X1,...,Xn] coincides with 0 that of w . Hence the valuation wσ on L[X1,...,Xn] extends both v0 and w; to simplify notation, we call it again v. By assumption, v is not symmetric on L[X1,...,Xm]. From Lemma 3.1 it follows that the set {vτπ | π ∈ Sn} of valuations on L[X1,...,Xn] has at least k + 1 = [L : K] + 1 many elements. But their restrictions INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 11 to K[X1,...,Xn] are wτπ, where τπ is now understood to be an automorphism of K[X1,...,Xn]. As w was chosen to be symmetric on K[X1,...,Xn], all of these restrictions coincide with w. Therefore, we have at least [L : K] + 1 many extensions of w to L[X1,...,Xn]. On the other hand, as the extension L[X1,...,Xn]|K[X1,...,Xn] is algebraic of degree [L : K], all of the extensions must be conjugate over K[X1,...,Xn] and there can be no more than [L : K] many. This contradiction shows that the valuation v must be symmetric on ac K [X1,...,Xm].

4. Proof of Theorem 1.3 and its corollaries ac Take an arbitrary field K and a valuation v on K(X1,...,Xm) . For m = 1 there is nothing to show, so we assume that m ≥ 2. ac Let us first show that if for every f ∈ K [X1,...,Xm] conditions (4) and (5) hold, then v is openly-symmetric. The proof will not depend on m; so if we just show that v is symmetric on K[X1,...,Xm] it will also follow that an extension of v which satisfies conditions (4) and (5) on K[X1,...,Xn] (with n in place of m), where n > m, is also symmetric. Taking condition (4) for the definition of such an extension, we then have proved that v is even openly-symmetric. Thus it actually suffices to show that v is symmetric. What we need for the proof of the symmetry are the following equations:

v(Xj − Xi) = δ for all distinct i, j ∈ {1, . . . , m},(14) ac (15) v(Xj − a) = v(X1 − a) for all a ∈ K and 2 ≤ j ≤ m . In order to obtain (14), we only need that condition (4) holds for polynomials in K[X1,...,Xm] because then, for every choice of distinct i, j ∈ {2, . . . , m},

v(Xj − Xi) = v(Xj − X1 + X1 − Xi)

= min{v(Xj − X1), v(X1 − Xi)} = δ . There are various conditions that ensure the validity of (15). In the case of the assumptions of our theorem, it works as follows: for 2 ≤ j ≤ m and every a ∈ Kac, v(Xj − a) = v(Xj − X1 + X1 − a) = min{v(Xj − X1), v(X1 − a)}

= v(X1 − a) where the second equality follows from (4) and the third equality follows from (5). 12 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

If we assume condition (4) only for polynomials in K[X1,...,Xm], as is the case in Corollaries 1.4 and 1.5, then we can proceed as follows. Suppose that condition (6) holds, but v(Xj − a) 6= v(X1 − a) for some ac a ∈ K . If v(Xj − a) < v(X1 − a), then

δ = v(Xj−X1) = min{v(Xj−a), v(X1−a)} = v(Xj−a) < v(X1−a) ≤ δ, where the ﬁrst equality holds by (14). This is a contradiction. If on the other hand v(Xj − a) > v(X1 − a), then the same argument with 1 and j interchanged also leads to a contradiction. This proves that (15) holds in this case. Now suppose that condition (7) holds. Then we obtain (15) because v(Xj − X1) = δ > v(X1 − a) implies that

v(Xj − a) = min{v(Xj − X1), v(X1 − a)} = v(X1 − a) .

Now we proceed to the proof of the symmetry. Take a polynomial g ∈ K[X] and write it as deg g Y g = c (X − ak) k=1 ac with c, ak ∈ K . Then for 2 ≤ j ≤ m, using (15), deg g X vg(X1) = vc + v(X1 − ak) k=1 deg g X = vc + v(Xj − ak) = vg(Xj) . k=1

Now take any polynomial f ∈ K[X1,...,Xm] and any π ∈ Sm . Write f in the form (3). Using the equalities we have computed above together with (14), we ﬁnd: vf = min vgi(X1) + (i2 + ... + im)δ i

= min vgi(Xπ(1)) + i2v(Xπ(2) − Xπ(1)) + ... + imv(Xπ(m) − Xπ(1)) i

i2 im = min vgi(Xπ(1))(Xπ(2) − Xπ(1)) · ... · (Xπ(m) − Xπ(1)) i

X i2 im ≤ v gi(Xπ(1))(Xπ(2) − Xπ(1)) · ... · (Xπ(m) − Xπ(1)) i

= vf(Xπ(1),...,Xπ(m)) = vτπf . Since f and π were arbitrary, we also have that

vτπf ≤ vτπ−1 (τπf) = v(τπ−1 τπf) = vf . INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 13

Altogether, this shows that vf = vτπf. We have thus shown that v is symmetric. This proves the corresponding implications in part a) of Theorem 1.3 and in Corollary 1.4, as well as Corollary 1.5.

Now we prove that if v is openly-symmetric on K[X1,...,Xm], then v must be as described in Theorem 1.3 and Corollary 1.4. From Propo- ac sition 1.1 we know that v is also openly-symmetric on K [X1,...,Xm]. First of all, because δ := v(X2 −X1) is invariant under permutations of the variables, all values v(Xj − Xi) are equal to δ, provided that ac i 6= j. Suppose that δ < v(X1 − a) for some a ∈ K . Then

v(X2 − a) = min{v(X2 − X1), v(X1 − a)} = δ < v(X1 − a) , ac contradicting the symmetry. Hence, δ ≥ v(X1 − a) for all a ∈ K , showing that (5) holds. Since v(Xi − a) = v(X1 − a) by symmetry, also (6) holds. ac Choose any d ∈ K(X1,X2) with vd = δ. For instance, d = X2 −X1 is a possible choice, but later on we will also want to consider other Xj −X1 choices. Then all elements d , 2 ≤ j ≤ m, are units in the valuation ring. ac Using that v is openly-symmetric on K [X1,...,Xm], extend it to an ac openly-symmetric valuation on K [X1,...,Xm+1]. Extend v further ac to a valuation on K(X1,...,Xm+1) . Remark 1.2 shows that v is ac symmetric on K(X1,...,Xi−1) [Xi,...,Xm+1] for 2 ≤ i ≤ m. Note that we are now in the situation where the assumptions of Corollary 1.6 are satisﬁed. The same arguments as before show that

ac (16) v(Xi − X1) = δ ≥ v(Xi − a) for all a ∈ K(X1,...,Xi−1) .

Xi−X1 For 3 ≤ i ≤ m we have that v d = 0, hence Lemma 2.4, (16) implies Xi−X1 that the residue of d is transcendental over the algebraically closed residue ﬁeld ac ac K(X1,...,Xi−1) v = (K(X1,...,Xi−1)v) .

X3−X1 Xm−X1 This shows that the residues of the elements d ,..., d are alge- ac braically independent over K(X1,X2) v, hence also over K(X1,X2)v. Taking d = X2 − X1, this proves part b) of the theorem. We continue with the proof of part a). ac If δ ∈ vK(X1) , then we can take the above arguments one step ac further. In this case we can take d ∈ K(X1) . Similarly as before, X2−X1 X2−X1 we obtain that v d = 0 and that d v is transcendental over ac K(X1) v, hence also over K(X1)v. We now have that the residues 14 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

X2−X1 Xm−X1 of the elements d ,..., d are algebraically independent over K(X1)v. ac ac Take any f ∈ K [X1,...,Xm]. If δ ∈ vK(X1) , then we take ac d ∈ K(X1) and rewrite the representation (3) as follows: f =

X i2 im = gi(X1) · (X2 − X1) · ... · (Xm − X1) i i2 im X X2 − X1 Xm − X1 = g (X ) · di2+...+im · · ... · . i 1 d d i

ac Using Lemma 2.1, taking K(X1) in place of K, I = ∅ and J = Xj −X1 {2, . . . , n} with yj = d , we obtain that vf = X − X i2 X − X im i2+...+im 2 1 m 1 = min vgi(X1) · d · · ... · i d d

= min (vgi(X1) + (i2 + ... + im)δ) , i which proves (4) in this case. ac If δ∈ / vK(X1) , then we take d = X2 − X1 and rewrite the representation (3) as follows: f =

X i2 im = gi(X1)(X2 − X1) ··· (Xm − X1) i i3 im X X3 − X1 Xm − X1 = g (X )(X − X )i2+...+im ··· . i 1 2 1 X − X X − X i 2 1 2 1

ac Using Lemma 2.1, taking K(X1) in place of K, I = {2} with x2 = Xj −X1 X2 − X1, and J = {3, . . . , m} with yj = v, we obtain that X2−X1 vf = X − X i3 X − X im i2+...+im 3 1 m 1 = min vgi(X1)(X2 − X1) ··· i X2 − X1 X2 − X1

= min (vgi(X1) + (i2 + ... + im)δ) , i which proves (4) in this second case. This completes the proof of part a) of Theorem 1.3, and of Corollary 1.4 and Corollary 1.6. INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 15

We turn to the proof of part c) of Theorem 1.3. Lemma 2.1 shows that in the ﬁrst of the above cases,

rr vK(X1,...,Xm)/vK(X1) = ac ac = rr vK(X1) (X2,...,Xm)/vK(X1) = 0 ,

trdeg K(X1,...,Xm)v|K(X1)v = ac ac = trdeg K(X1) (X2,...,Xm)v|K(X1) v = m − 1 , and in the second case,

rr vK(X1,...,Xm)/vK(X1) = ac ac = rr vK(X1) (X2,...,Xm)/vK(X1) = 1 ,

trdeg K(X1,...,Xm)v|K(X1)v ac ac = trdeg K(X1) (X2,...,Xm)v|K(X1) v = m − 2 . Both cases obviously appear as we are free to choose δ as long as it satisﬁes condition (5).

We have seen that the valuation on K(X1) can be arbitrary. By Lemma 2.1 there are always extensions of v from K to K(X1) such that K(X1)v|Kv = 1, which by the Abhyankar inequality (12) forces rr vK(X1)/vK = 0. Combining such a valuation with the two cases above, we obtain openly-symmetric valuations on K(X1,...,Xm) such that

rr vK(X1,...,Xm)/vK = 0 and trdeg K(X1,...,Xm)v|Kv = m in the first case, and rr vK(X1,...,Xm)/vK = 1 and trdeg K(X1,...,Xm)v|Kv = m − 1 in the second case. This shows that cases (a) and (c) of Theorem 1.3 can appear. Likewise, Lemma 2.1 shows that there are always extensions of v from K to K(X1) such that rr vK(X1)/vK = 1, which by the Abhyankar inequality (12) forces K(X1)v|Kv = 0. Combining such a valuation with the two cases above, we obtain openly-symmetric valuations on K(X1,...,Xm) such that rr vK(X1,...,Xm)/vK = 1 and trdeg K(X1,...,Xm)v|Kv = m − 1 in the first case, and rr vK(X1,...,Xm)/vK = 2 and trdeg K(X1,...,Xm)v|Kv = m − 2 in the second case. This shows that, again, case (c) can appear, and that case (e) can appear. Note that we have realized case (c) in two different ways. 16 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

If the valuation v on K admits an extension to K(X1) that does not enlarge value group and residue ﬁeld, then we get rr vK(X1)/vK = 0 and trdeg K(X1)v|Kv = 0. Combining this with the two cases above, we obtain openly-symmetric valuations on K(X1,...,Xm) such that rr vK(X1,...,Xm)/vK = 0 and trdeg K(X1,...,Xm)v|Kv = m − 1 in the ﬁrst case, and rr vK(X1,...,Xm)/vK = 1 and trdeg K(X1,...,Xm)v|Kv = m − 2 in the second case. Thus cases (b) and (d) can appear. This completes the proof of Theorem 1.3.

It remains to prove Corollary 1.7. Assume that the valuation vδ is defined on K(X1,...,Xm) by definition (8). We can extend vδ to ac K (X1,...,Xm) by applying the same definition to all polynomials f ∈ ac K (X1,...,Xm). Now it suffices to prove that vδ is openly-symmetric ac on K (X1,...,Xm). ac From our definition of vδ on K (X1,...,Xm), we obtain that for all a ∈ Kac,

(17) δ ≥ min{δ, va} = v(X1 − a) . Furthermore, we obtain that for 2 ≤ i ≤ m,

(18) vδ(Xi − X1) = min{δ, δ} = δ . ac We consider the valuation v deﬁned by (4) on K (X1,...,Xm), ac where we take v = vδ on K (X1). From (17) it follows that condition (5) is satisﬁed, so we know from Theorem 1.3 that v is openly- symmetric. Therefore, we obtain that for 2 ≤ i ≤ m,

vXi = vX1 = vδX1 = δ . ac We show that vδ coincides with v on K (X1,...,Xm). Indeed, for ac every f ∈ K [X1,...,Xm] written in the form (1),

i1 im vδf = min (vdi + (i1 + ... + im)δ) = min v diX1 · ... · Xm i i

X i1 im ≤ v diX1 · ... · Xm = vf , i and if we write f in the form (3), then by deﬁnition of v and by (18),

vf = min (vgi(X1) + (i2 + ... + im)δ) i

i2 im = min vδ gi(X1)(X2 − X1) · ... · (Xm − X1) i

X i2 im ≤ vδ gi(X1)(X2 − X1) · ... · (Xm − X1) = vδf . i INVARIANT VALUATIONS ON RATIONAL FUNCTION FIELDS 17

ac It follows that vδ = v on K (X1,...,Xm), which proves that vδ is openly-symmetric. Finally, let us note that the comparison method of the above proof can be adapted to show the following fact, which is of independent interest. Corollary 4.1. Take a valued ﬁeld (K, v) and two valuations w, w0 extending v from K to a rational function ﬁeld

K(X1,...,Xm) = K(Y1,...,Ym) , induced by the deﬁnitions

X i1 im w diX1 · ... · Xm := min (vdi + i1δ1 + ... + imδm) , i i

0 X j1 jm w djY1 · ... · Ym := min (vdj + j1δ1 + ... + jmδm) . j j Then w = w0.

5. An example We construct an example of a valuation v on the rational function ac ac ac field Q (X,Y ) over Q which is not symmetric on Q (X,Y ) but√ has a symmetric restriction to Q(X,Y ). To this end, we write i = −1 and extend the trivial valuation on Qac(X) to a valuation v on the rational function field Qac(X,Y ) = Qac(X,Y − iX) by definition (8), ac where we take K = Q (X), δ = 1, m = 1 and X1 = Y − iX. So we have v(Y − iX) = 1 . Note that (19) v(X + iY ) = vi + v(Y − iX) = v(Y − iX) = 1 , (20) v(Y + iX) = min{v(Y − iX), v(2iX)} = 0 , (21) v(X2 + Y 2) = v(Y − iX) + v(Y + iX) = 1 . Equations (19) and (20) together show that v is not symmetric on Qac(X,Y ), and even not on Q(i)(X,Y ). We show that v is not only trivial on Qac(X), but also on Qac(Y ). It suffices to show that v is trivial on Qac[Y ], and since it is trivial on Qac and every polynomial in Qac[Y ] splits into linear factors, it even suffices to show that v(Y − a) = 0 for every a ∈ Qac. Using that iX − a 6= 0 since iX∈ / Qac and that v is trivial on Qac(X), we compute: v(Y − a) = min{v(Y − iX), v(iX − a)} = 0 . 18 F.-V. KUHLMANN, K. KUHLMANN, AND C. VIS¸AN

Now we show that the value of every nonzero f ∈ Q(X)[Y ] with degY f ≤ 1 is 0. This is clear when the degree is 0 since then f ∈ ac Q(X) ⊂ Q (X), on which v is trivial. For arbitrary g0, g1 ∈ Q(X) with g1 6= 0, we compute: g0 g0 v(g1Y + g0) = vg1 + v Y + = v Y + g1 g1 g = min v(Y − iX), v iX + 0 = 0 , g1 where we use that iX + g0/g1 6= 0 since −iX∈ / Q(X). Similarly, we show that the value of every f ∈ Q(Y )[X] with degX f ≤ 1 is 0. This is clear when the degree is 0 since then f ∈ Q(Y ) ⊂ Qac(Y ), on which v is trivial. For arbitrary h0, h1 ∈ Q(Y ) with h1 6= 0, we compute: h0 h0 v(h1X + h0) = vh1 + v X + = v X + h1 h1 h = min v(X + iY ), v −iY + 0 = 0 , h1 where we use (19) and that −iY + h0/h1 6= 0 since iY∈ / Q(Y ). Set F (Y ) = X2 + Y 2 ∈ Q(X)[Y ]. Using the F -adic expansion in Q(X)[Y ], every polynomial in f ∈ Q[X,Y ] can be written in the form X 2 2 i f(X,Y ) = fi(X,Y )(X + Y ) i with fi(X,Y ) ∈ Q(X)[Y ] such that degY fi(X,Y ) ≤ 1. Then also degX fi(Y,X) ≤ 1, and by what we have shown before,

vfi(X,Y ) = 0 = vfi(Y,X) for all i such that fi(X,Y ) 6= 0. In view of (21), we obtain:

X 2 2 i vf(X,Y ) = v fi(X,Y )(X + Y ) = i

= min{i | fi(X,Y ) 6= 0} = min{i | fi(Y,X) 6= 0} X 2 2 i = v fi(Y,X)(Y + X ) = vf(Y,X) , i which shows that v is symmetric on Q(X,Y ).

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Institute of Mathematics, University of Silesia, Bankowa 14, 40-007 Katowice, Poland E-mail address: [email protected]

Romanian-American University of Bucharest, 1B Expozitiei Blv., Sector 1, Bucharest, Romania E-mail address: [email protected]